JEE Main 2022 · 28 June · evening · Single correct · +4 / −1
The correct IUPAC name of the following compound is :
- (a) 4-methyl-2-nitro-5-oxohept-3-enal
- (b) 4-methyl-5-oxo-2-nitrohept-3-enal
- (c) 4-methyl-6-nitro-3-oxohept-4-enal
- (d) 6-formyl-4-methyl-2-nitrohex-3-enal
Answer and solution
Answer (c) 4-methyl-6-nitro-3-oxohept-4-enal
Solution
Chemistry ma’am · 4 steps
The aldehyde is the senior group, so its carbon is C1. Number the whole chain from that end.
- Principal group. The molecule has -CHO, a ketone C=O and -NO₂. The aldehyde ranks highest, so the name ends in -al. Its carbon must be C1 of the main chain.
- Number the chain. Start at CHO (C1). Then CH₂ is C2, the ketone carbon is C3, the carbon carrying CH₃ is C4, =CH⁻ is C5, CH(NO₂) is C6 and the end CH₃ is C7. That gives 7 carbons with the double bond at C4, so the parent is hept-4-enal.
Carbon Group on it Prefix C3 C=O (ketone) 3-oxo C4 -CH₃ 4-methyl C6 -NO₂ 6-nitro - Alphabetical order. Prefixes go in alphabetical order: methyl, nitro, oxo. Put them before the parent name.
- So. The name is 4-methyl-6-nitro-3-oxohept-4-enal, option (c).
The idea to keep. Find the highest-priority group first and give it the lowest number. For an aldehyde, its carbon is always C1. Everything else follows from that start, and prefixes are listed alphabetically.
Answer from the NTA final answer key. Solution by PrepWiser.